package algorithm.problems.hashing;

import java.util.*;

/**
 * Created by gouthamvidyapradhan on 01/05/2018.
 * S and T are strings composed of lowercase letters. In S, no letter occurs more than once.

 S was sorted in some custom order previously. We want to permute the characters of T so that they match the order
 that S was sorted. More specifically, if x occurs before y in S, then x should occur before y in the returned string.

 Return any permutation of T (as a string) that satisfies this property.

 Example :
 Input:
 S = "cba"
 T = "abcd"
 Output: "cbad"
 Explanation:
 "a", "b", "c" appear in S, so the order of "a", "b", "c" should be "c", "b", and "a".
 Since "d" does not appear in S, it can be at any position in T. "dcba", "cdba", "cbda" are also valid outputs.


 Note:

 S has length at most 26, and no character is repeated in S.
 T has length at most 200.
 S and T consist of lowercase letters only.

 Solution: O(N) count occurrence of each character and write to the output string
 */
public class CustomSortString {

    /**
     * Main method
     * @param args
     * @throws Exception
     */
    public static void main(String[] args) throws Exception{
        System.out.println(new CustomSortString().customSortString("cba", "abcd"));
    }

    public String customSortString(String S, String T) {
        Map<Character, Integer> map = new HashMap<>();
        for(int i = 0; i < T.length(); i ++){
            if(!map.containsKey(T.charAt(i))){
                map.put(T.charAt(i), 1);
            } else{
                map.put(T.charAt(i), map.get(T.charAt(i)) + 1);
            }
        }
        StringBuilder result = new StringBuilder();
        for(char c : S.toCharArray()){
            if(map.containsKey(c)){
                int count = map.remove(c);
                for(int i = 0; i < count; i ++){
                    result.append(c);
                }
            }
        }
        for(char c : map.keySet()){
            int count = map.get(c);
            for(int i = 0; i < count; i ++){
                result.append(c);
            }
        }
        return result.toString();
    }
}
